Answer:
Explanation:
The strong bases have following properties:
1. In solution, strong bases ionize fully.
2. On dissolving the strong bases in water they produce all hydroxide ion which they have.
3. For strong bases the value of equilibrium constant (Kb ) is large.
4. In general the strong base ionizes completely means concentration of ions are greater means conductivity also greater.
5. For strong bases the value of equilibrium constant (Kb) is large, thus the value of dG0 is very large negative number.
Is bubble chamber one of your choices? Bubble chamber sounds like a good fit for the question.
Answer:
The final mass of sample is 1.3 g.
Explanation:
Given data:
Half life of H-3 = 12.32 years
Amount left for 15.0 years = 3.02 g
Final amount = ?
Solution:
First all we will calculate the decay constant.
t₁/₂ = ln² /k
t₁/₂ =12.32 years
12.32 y = ln² /k
k = ln²/12.32 y
k = 0.05626 y⁻¹
Now we will find the original amount:
ln (A°/A) = Kt
ln (3.02 g/ A) = 0.05626 y⁻¹ × 15.0 y
ln (3.02 g/ A) = 0.8439
3.02 g/ A = e⁰°⁸⁴³⁹
3.02 g/ A = 2.33
A = 3.02 g/ 2.33
A = 1.3 g
The final mass of sample is 1.3 g.
Answer:
ΔS = -661.0J/mol is the entropy change for the system
ΔS = -842J/mol.K is the entropy change for the surroundings
Explanation:
From the relationship between ΔG, T, ΔH and ΔS,
Mathematically, ΔG = ΔH - TΔS
TΔS = ΔH - ΔS
ΔS = ΔH - ΔS / T
but ΔG = -54 kJ/mol, ΔH = -251 kJ/mol and T = 25 °C (298 K)
plugging into the equation,
ΔS = -251 kJ/mol - ( -54 kJ/mol) / 298
ΔS = -0.6610KJ/mol or in J.mol
ΔS = -661.0J/mol is the entropy change for the system
- For entropy change for the surroundings = ΔS = ΔH/T
- ΔS = -0.84KJ/mol.K or -842J/mol.K is the entropy change for the surroundings
